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Nat2105 [25]
3 years ago
14

PLEASE HELP!! I HAVE TO TURN THIS IN IN 10 MINUTES

Mathematics
1 answer:
grin007 [14]3 years ago
6 0

Answer:

0.8

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer/how I got this answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

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A particular baseball player has a batting average (number of hits/number of times at bat) of 0.364. What is the probability tha
MAXImum [283]
The correct answer is option (c) <span>P(not a hit) = 0.636

</span>The batting average of the baseball player = \frac{number of hits}{number of times at bat} = 0.364

Now, the probability of occurrence of any event is say, P, the probability of not occurrence of the event is given as =  1 - P

Therefore, <span>the probability that he will not get a hit at his next at bat = 1- 0.364 = 0.636</span>
6 0
3 years ago
Two cards are selected from a standard deck of 52 playing cards. the first is replaced before the second card is selected. find
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<span>The correct answer is
</span>  Find the probability of selecting a diamond and then selecting a queen.

 standard deck of 52 playing cards

<span>hearts: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards</span>

<span>clubs: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards</span>

<span>spades: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards</span>

<span>diamonds: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A---- > 13 cards</span>

 a)  The probability of selecting a diamond is 13/52

<span>b)  </span>The probability of selecting a queen is 4/52

<span>c)   </span>The probability of selecting a diamond and then selecting a queen is

(13/52)*(4/52)=0.0192------------->1.92%

6 0
3 years ago
The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr
aniked [119]

Answer:

(a) E(X) =  0.383

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) P(x = 4) = 0.000169

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$ E(X) =  \frac{n \times r}{N} $

$ E(X) =  \frac{10 \times 8}{209} $

$ E(X) =  \frac{80}{209} $

E(X) =  0.383

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$ P(x = 4) =  \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}} $

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$ P(x = 4) =  \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{70 \times 84944276340}{35216131179263320}

P(x = 4) = 0.000169

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0
3 years ago
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Step-by-step explanation: Hope its fine.

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Answer:

D.No solution

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